The Final Angular Speed of a Rotating Cylinder System

What is the final angular speed of the system consisting of a rotating cylinder and a piece of putty?

Given data:

Mass of the cylinder (M): 12.2 kg

Radius of the cylinder (R): 1.00 m

Angular speed of the cylinder (ω): 5.50 rad/s

Mass of the putty (m): 0.250 kg

Distance of the putty from the center (r): 0.900 m

Answer:

The final angular speed of the system that contains a solid, horizontal cylinder of mass 12.2 kg and radius 1.00 m rotating with an angular speed of 5.50 rad/s about a fixed vertical axis through its center is 5.33 rad/s.

When a 0.250-kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m from the center of rotation and sticks to the cylinder, the final angular speed of the system can be determined using the conservation of angular momentum.

According to the law of conservation of angular momentum:

1/2 * M * R² * ω = (1/2 * M * R² + m * r²) * ω'

Substitute the given values into the equation:

1/2 * 12.2 * 1 * 5.5 = [(1/2 * 12.2 * 1) + (0.25 * 0.9²)] * ω'

ω' = 33.55 / 6.3

ω' = 5.33 rad/s

Therefore, the final angular speed of the system is 5.33 rad/s after the piece of putty is dropped onto the rotating cylinder and sticks to it.